Problem: The equation of a circle $C$ is $x^2+y^2+14x-12y+36 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+14x) + (y^2-12y) = -36$ $(x^2+14x+49) + (y^2-12y+36) = -36 + 49 + 36$ $(x+7)^{2} + (y-6)^{2} = 49 = 7^2$ Thus, $(h, k) = (-7, 6)$ and $r = 7$.